On Index Divisors and Monogenity of Certain Number Fields Defined by Trinomials x7+ax+b

Lhoussain El Fadil

doi:10.20944/preprints202309.0481.v1

Submitted:

05 September 2023

Posted:

07 September 2023

You are already at the latest version

Abstract

In this paper for every number field K generated by a root α of a trinomial x⁷ + ax + b ∈ Z[x] and for every prime integer p, we calculate ν_p(i(K)), the highest power of p dividing the index i(K) of the field K. In particular, we calculate the index i(K). As application, when the index of K is not trivial, then K is not monogenic.

Keywords:

Power integral bases

;

theorem of Ore

;

prime ideal factorization

;

common index divisor

Subject:

Computer Science and Mathematics - Algebra and Number Theory

MSC: 11R04; 11Y40; 11R21

1. Introduction

Let

K

be a number field of degree

n

,

Z_{K}

its ring of integers, and

d_{K}

its absolute discriminant. It is well known that

Z_{K}

is a free abelian group of rank

n

and by the fundamental theorem of finite abelian groups,

(Z_{K} : Z [θ])

is a finite group for every primitive element

θ \in Z_{K}

of

K

. Let

i n d (θ) = (Z_{K} : Z [θ])

.

i n d (θ)

is called the index of

θ

. The index of the number field

K

is defined by

i (K) = G C D ((Z_{K} : Z [θ]) ∣ K = Q (θ) a n d θ \in Z_{K})

. A rational prime integer

p

dividing

i (K)

is called a prime common index divisor of

K

. The number field

K

is called monogenic if it admits a

Z

basis of type

(1, θ, \dots, θ^{n - 1})

for some

θ \in Z_{K}

. Remark that if

Z_{K}

has a power integral basis, then

i (K) = 1

. Therefore a field having a prime common index divisor is not monogenic. Monogenity of number fields is a classical problem of algebraic number theory, going back to Dedekind, Hasse and Hensel, see for instance [19,25,26] for the present state of this area. It is called a problem of Hasse to give an arithmetic characterization of those number fields which are monogenic [23,25,26]. For any primitive element

θ \in Z_{K}

of

K

, it is well-known that

| ▵ (θ) | = i n d {(θ)}^{2} \cdot | d_{K} |

where

▵ (θ)

is the discriminant of the minimal polynomial of

θ

over

Q

[19].

Clearly,

i n d (θ) = 1

for some primitive element

θ \in Z_{K}

of

K

if and only if

(1, θ, \dots, θ^{n - 1})

is a power integral basis of

Z_{K}

.

The problem of testing the monogenity of number fields and constructing power integral bases have been intensively studied during the last four decades mainly by Gaál, Györy, Nakahara, Pohst and their collaborators (see for instance [1,16,32]). In 1871, Dedekind was the firstone who gave an example of a number field with non trivial index, he considered the cubic field

K

generated by a root of

x^{3} - x^{2} - 2 x - 8

and showed that the rational prime

2

splits completely in

K

([5, § 5, page 30]). According to a well known theorem of Dedekind ([24, Chapter I, Proposition 8.3]), if we suppose that

K

is monogenic, then we would be able to find a cubic polynomial defining

K

, that splits completely into distinct polynomials of degree

1

in

F_{2} [x]

. Since there is only two distinct polynomials of degree

1

in

F_{2} [x]

, this is impossible. In

1930

, Engstrom was the first one who related the prime ideal factorization and the index of a number field of degree less than

8

[13]. For any number field

K

of degree

n \leq 7

, he showed that

ν_{p} (i (K))

is explicitly determined by the factorization of

p Z_{K}

into powers of prime ideals of

K

for every positive rational prime integer

p \leq n

. This motivated Narkiewicz to ask a very important question, stated as problem 22 in Narkiewicz’s book ([31, Problem 22]), which asks for an explicit formula of the highest power

ν_{p} (i (K))

for a given rational prime

p

dividing

i (K)

. In [30], Nakahara studied the index of non-cyclic but abelian biquadratic number fields. He showed that the field index of such fields is in the set

{1, 2, 3, 4, 6, 12}

. In [17] Gaál et al. characterized the field indices of biquadratic number fields having Galois group

V_{4}

and they proved that

i (K) \in {1, 2, 3, 4, 6, 12}

. Recently, many authors are interested on monogenity of number fields defined by trinomials. Davis and Spearman [6] studied the index of quartic number fields

K

generated by a root of such a quartic trinomial

F (x) = x^{4} + a x + b \in Z [x]

. They gave necessary and sufficient conditions on

a

and

b

so that a prime

p

is a common index divisor of

K

for

p = 2, 3

. Their method is based on the calculation of the

p

-index form of

K

, using

p

-integral bases of

K

. El Fadil and Gaál [11] studied the index of quartic number fields

K

generated by a root of a quadratic trinomial of the form

F (x) = x^{4} + a x^{2} + b \in Z [x]

. They gave necessary and sufficient conditions on

a

and

b

so that a prime

p

is a common index divisor of

K

for every prime integer

p

. In [15], for a sextic number field

K

defined by a trinomial

F (x) = x^{6} + a x^{3} + b \in Z [x]

, Gaál studied the multi-monegenity of

K

; he calculated all possible power integral bases of

K

. In [9], we extended Gaál’s studies by providing some cases where

K

is not monogenic. Also in [10], for every prime integer

p

, we gave necessary and sufficient conditions on

a

and

b

so that

p

is a common index divisor of

K

, where

K

is a number field defined by an irreducible trinomial

F (x) = x^{5} + a x^{2} + b \in Z [x]

. In [8], we provided some sufficient conditions which guarantee that

i (K)

is not trivial, and so

K

is not mongenic. In this paper, for a septic number field generated by a root of a trinomial

F (x) = x^{7} + a x + b \in Z [x]

and for every prime integer

p

, we calculate

ν_{p} (i (K))

, the highest power of

p

dividing the index

i (K)

of the field

K

. Our method is based on Newton’s polygon techniques applied in prime ideal factorization, which is performed in [21,22] and in Montes’ thesis defended in 1999. The author is very thankful to Professor Enric Nart who provided him a copy of Montes’ thesis.

2. Main Results

Throughout this section

K

is a number field generated by a root

α

of an irreducible trinomial

F (x) = x^{7} + a x + b \in Z [x]

and we assume that for every rational prime integer

p

,

ν_{p} (a) \leq 5

or

ν_{p} (b) \leq 6

. Along this paper, for every integer

a \in Z

and a prime integer

p

, let

a_{p} = \frac{a}{p^{ν_{p} (a)}}

.

We start with the following theorem, which characterizes when is

Z [α]

integrally closed?

Theorem 2.1.

The ring

Z [α]

is integrally closed if and only if every prime integer

p

satisfies one of these conditions:

1.: If $p | a$ and $p | b$ , then $ν_{p} (b) = 1$ .
2.: If $p = 2$ , $p$ divides $b$ and does not $a$ , then $(a, b) \in {(1, 0), (3, 2)} (\mod 4)$ .
3.: If $p = 3$ , $p$ divides $b$ and $a \equiv - 1 (\mod 3)$ , then $(a, b) \in {(2, 0), (8, 3), (8, 6)} (\mod 9)$ .
4.: If $p = 3$ , $p$ divides $b$ and $a \equiv 1 (\mod 3)$ , then $(a, b) \neg \equiv (1, 0) (\mod 9)$ .
5.: If $p = 7$ , $p$ divides $a$ and does not divide $b$ , then $ν_{7} (1 - a - b^{6}) = 1$ .
6.: If $p \notin {2, 3, 7}$ , $p$ does not divide both $a$ and $b$ , then $ν_{p} (7^{7} b^{6} + 6^{6} a^{7}) \leq 1$ .

The following example gives an infinite family of monogenic septic number fields defined by non monogenic trinomials.

Proposition 2.2.

Let

K

be the number field generated by a root

α

of

F (x) = x^{7} + 2^{u} a x + 2^{v} b \in Z [x]

, with

u \geq v - 1

,

2 \leq v \leq 6

, GCD

(6, b) = 1

,

7

does not divide

a

, and for every odd prime integer

p

, if

p

does not divide

b

, then

p^{2}

does not divide

7^{7} b^{6} + 6^{6} a^{7}

. Then

F (x)

is a non monogenic polynomial and

K

is a monogenic number field.

In the remainder of this section, for every prime integer

p

and for every values of

a

and

b

, we calculate

ν_{p} (i (K))

. For every integers

a

and

b

, let

▵ = - (6^{6} a^{7} + 7^{7} b^{6})

be the discriminant of

F (x)

and for every prime integer

p

, let

▵_{p} = \frac{▵}{p^{ν_{p} (▵)}}

.

Theorem 2.3.

The following table provides the value of

ν_{2} (i (K))

.

\begin{matrix} c o n d i t i o n s & ν_{2} (i (K)) \\ a \equiv 28 (\mod 32) a n d b \equiv 0 (\mod 32) & 1 \\ a \equiv 112 (\mod 128) a n d b \equiv 0 (\mod 128) & 1 \\ a \equiv 1 (\mod 8) a n d b \equiv 2 (\mod 4) \\ ν_{2} (▵) e v e n a n d ▵_{2} \equiv 3 (\mod 4) & 1 \\ a \equiv 3 (\mod 8) a n d b \equiv 4 (\mod 8) & 1 \\ a \equiv 3 (\mod 4) a n d b \equiv 0 (\mod 8) & 3 \\ (a, b) \in {(5, 2), (5, 6), (13, 2), (13, 14)} (\mod 16) & 1 \\ O t h e r w i s e & 0 \end{matrix}

Theorem 2.4.

The following table provides the value of

ν_{3} (i (K))

.

\begin{matrix} c o n d i t i o n s & ν_{3} (i (K)) \\ a \equiv 5 (\mod 9) a n d b \in {3, 6} (\mod 9) & 1 \\ a \equiv 8 (\mod 9) a n d b \equiv 0 (\mod 9) & 2 \\ a \equiv 2 (\mod 9) a n d b \in {3, 6} (\mod 9) \\ ν_{3} (▵) = 2 k a n d k \geq 5 & 1 \\ a \equiv 2 (\mod 9) a n d b \in {3, 6} (\mod 9) \\ ν_{3} (▵) = 2 k + 1, k \geq 5 a n d ▵_{3} \equiv 1 (\mod 3) & 2 \\ O t h e r w i s e & 0 \end{matrix}

Theorem 2.5.

For every prime integer

p \geq 5

and for every integers

a

and

b

such that

F (x) = x^{7} + a x + b

is irreducible over

Q

,

p

does not divide

i (K)

, where

K

is the number field defined by

F (x)

.

Corollary 2.6.

For every integers

a

and

b

such that

F (x) = x^{7} + a x + b

is irreducible over

Q

,

i (K) \in {1, 2, 3, 6, 8, 9, 18, 24, 72}

.

Remark 1.

1.: The field $K$ can be non monogenic even if the index $i (K) = 1$ .
2.: The unique method which allows to test whether $K$ is monogenic is to calculate the solutions of the index form equation of the field $K$ (see for instance [18,19]).

3. A short introduction to prime ideal factorization based on Newton polygons

In 1894, Hensel developed a powerful approach by showing that for every prime integer

p

, the prime ideals of

Z_{K}

lying above

p

are in one–one correspondence with monic irreducible factors of

F (x)

in

Q_{p} [x]

. For every prime ideal corresponding to any irreducible factor in

Q_{p} [x]

, the ramification index and the residue degree together are the same as those of the local field defined by the associated irreducible factor [28]. Since then, to factorize

p Z_{K}

, we need to factorize

F (x)

in

Q_{p} [x]

. Newton’s polygon techniques can be used to refine the factorization. This is a standard method which is rather technical but very efficient to apply. We have introduced the corresponding concepts in several former papers. Here we only give a brief introduction which makes our proofs understandable. For a detailed description, we refer to Ore’s Paper [34] and Guardia, Montes and Nart’s paper [20]. For every prime integer

p

, let

ν_{p}

be the

p

-adic valuation of

Q_{p}

and

Z_{p}

the ring of

p

-adic integers. Let

F (x) \in Z_{p} [x]

be a monic polynomial and

ϕ \in Z_{p} [x]

a monic lift of an irreducible factor of

\bar{F (x)}

modulo

p

. Let

F (x) = a_{0} (x) + a_{1} (x) ϕ (x) + \dots + a_{n} (x) ϕ {(x)}^{l}

be the

ϕ

-expansion of

F (x)

,

N_{ϕ} (F)

the

ϕ

-Newton polygon of

F (x)

and

N_{ϕ}^{+} (F)

its principal part. Let

F_{ϕ}

be the field

F_{p} [x] / (\bar{ϕ})

. For every side

S

of

N_{ϕ}^{+} (F)

with length

l

and initial point

(s, u_{s})

, for every

i = 0, \dots, l

, let

c_{i} \in F_{ϕ}

be the residue coefficient, defined as follows:

c_{i} = \{\begin{matrix} 0, & if (s + i, u_{s + i}) lies strictly above S, \\ (\frac{a_{s + i} (x)}{p^{u_{s + i}}}) mod (p, ϕ (x)), & if (s + i, u_{s + i}) lies on S . \end{matrix}

Let

- λ = - h / e

be the slope of

S

, where

h

and

e

are two positive coprime integers. Then

d = l / e

is the degree of

S

. Let

R_{1} (F) (y) = t_{d} y^{d} + t_{d - 1} y^{d - 1} + \dots + t_{1} y + t_{0} \in F_{ϕ} [y]

, called the residual polynomial of

F (x)

associated to the side

S

, where for every

i = 0, \dots, d

,

t_{i} = c_{i e}

. If

R_{1} (F) (y)

is square free for each side of the polygon

N_{ϕ}^{+} (F)

, then we say that

F (x)

is

ϕ

-regular.

Let

\bar{F (x)} = \prod_{i = 1}^{r} {\bar{ϕ_{i}}}^{l_{i}}

be the factorization of

F (x)

into powers of monic irreducible coprime polynomials over

F_{p}

, we say that the polynomial

F (x)

is

p

-regular if

F (x)

is a

ϕ_{i}

-regular polynomial with respect to

p

for every

i = 1, \dots, r

. Let

N_{ϕ_{i}}^{+} (F) = S_{i 1} + \dots + S_{i r_{i}}

be the

ϕ_{i}

-principal Newton polygon of

F (x)

with respect to

p

. For every

j = 1, \dots, r_{i}

, let

R_{1_{i j}} (F) (y) = \prod_{s = 1}^{s_{i j}} ψ_{i j s}^{a_{i j s}} (y)

be the factorization of

R_{1_{i j}} (F) (y)

in

F_{ϕ_{i}} [y]

, where

R_{1_{i j}} (F) (y)

is the residual polynomial of

F (x)

attached to the side

S_{i j}

. Then we have the following theorem of index of Ore:

Theorem 3.1.

([12, Theorems 1.7 and 1.9])

Under the above hypothesis, we have the following:

1.: $ν_{p} ((Z_{K} : Z [α])) \geq \sum_{i = 1}^{r} {ind}_{ϕ_{i}} (F) .$

The equality holds if $F (x) is p$ -regular.
2.: If $F (x) is p$ -regular, then

$p Z_{K} = \prod_{i = 1}^{r} \prod_{j = 1}^{t_{i}} \prod_{s = 1}^{s_{i j}} p_{i j s}^{e_{i j}}$

is the factorization of $p Z_{K}$ into powers of prime ideals of $Z_{K}$ , where $e_{i j}$ is the smallest positive integer satisfying $e_{i j} λ_{i j} \in Z$ and the residue degree of $p_{i j s}$ over $p$ is given by $f_{i j s} = deg (ϕ_{i}) \times deg (ψ_{i j s})$ for every $(i, j, s)$ .

The Dedekind criterion can be reformulated as follows:

Theorem 3.2.

([7, Theorem 1.1])

Under the above hypothesis, let

R_{i} (x)

be the remainder of the Euclidean division of

F (x)

by

ϕ_{i} (x)

. Then

ν_{p} ((Z_{K} : Z [α])) = 0

if and only if

l_{i} = 1

or

ν_{p} (R_{i} (x)) = 1

for every

i = 1, \dots, r

.

When the theorem of Ore fails, that is

F (x)

is not

p

-regular, then in order to complete the factorization of

F (x)

, Guardia, Montes, and Nart introduced the notion of high order Newton polygon. By analogous to the first order, for each order

r

, the authors of [20] introduced the valuation

ω_{2}

of order

r

, the key polynomial

ϕ_{2} (x)

of such a valuation,

N_{r} (F)

the Newton polygon of any polynomial

F (x)

with respect to

ω_{2}

and

ϕ_{2} (x)

, and for every side of

N_{r} (F)

the residual polynomial

R_{r} (F)

, and the index of

F (x)

in order

r

. For more details, we refer to [20].

4. Proofs of our main results

Proof of Theorem 2.1.

If $p$ divides $a$ and $b$ , then by Theorem 3.2, $p$ does not divide $(Z_{K} : Z [α])$ if and only if $ν_{p} (b) = 1$ .
For $p = 2$ , $2$ divides $b$ and does not $a$ , we have $\bar{F (x)} = x {(x - 1)}^{2} {(x^{2} + x + 1)}^{2}$ . Let $ϕ_{1} = x - 1$ and $ϕ_{2} = x^{2} + x + 1$ . Since $F (x) = \dots - (4 x - 2) ϕ_{2} + (a + 1) x + b$ and $F (x) = \dots + (a + 7) ϕ_{1} + (a + 1 + b)$ , by Theorem 3.2, $2$ does not divide $(Z_{K} : Z [α])$ if and only if $ν_{2} (b + a + 1) = 1$ and $ν_{2} ((a + 1) x + b) = 1$ , which means $b \equiv 1 - a (\mod 4)$ and $a \equiv 1 (\mod 4)$ or $b \equiv 2 (\mod 4)$ . That is $(a, b) \in {(1, 0), (3, 2)} (\mod 4)$ .
For $p = 3$ , $3$ divides $b$ and $a \equiv 1 (\mod 3)$ , we have $\bar{F (x)} = x {(x^{2} + 1)}^{3}$ . Let $ϕ = x^{2} + 1$ . Since $F (x) = x ϕ^{3} - 3 x ϕ^{2} + 3 x ϕ + (a - 1) x + b$ , by Theorem 3.2, $3$ does not divide $(Z_{K} : Z [α])$ if and only if $ν_{3} ((a - 1) x + b) = 1$ , which means that $a \neg \equiv 1 (\mod 9)$ or $b \neg \equiv 0 (\mod 9)$ . That is $(a, b) \neg \equiv (1, 0) (\mod 9)$ .
For $p = 3$ , $3$ divides $b$ and $a \equiv - 1 (\mod 3)$ , we have $\bar{F (x)} = x {(x - 1)}^{3} {(x + 1)}^{2}$ . Let $ϕ_{1} = x - 1$ and $ϕ_{2} = x + 1$ . Since $F (x) = ϕ_{1}^{7} + 7 ϕ_{1}^{6} + 21 ϕ_{1}^{5} + 35 ϕ_{1}^{4} + 35 ϕ_{1}^{3} + 21 ϕ_{1}^{2} + (a + 7) ϕ_{1} + (a + 1 + b)$ and $F (x) = ϕ_{1}^{7} - 7 ϕ_{2}^{6} + 21 ϕ_{2}^{5} - 35 ϕ_{2}^{4} + 35 ϕ_{2}^{3} - 21 ϕ_{2}^{2} + (a + 7) ϕ_{2} + (b - a - 1)$ , by Theorem 3.2, $3$ does not divide $(Z_{K} : Z [α])$ if and only if $ν_{2} (a + 1 + b) = 1$ and $ν_{3} (b - a - 1) = 1$ . That is $(a, b) \in {(2, 0), (8, 3), (8, 6)} (\mod 9)$ .
For $p = 7$ , if $7$ divides $a$ and $7$ does not divide $b$ , then $\bar{F (x)} = {(x + b)}^{7}$ . Let $ϕ = x + b$ . Then $F (x) = ϕ^{7} - 7 b ϕ^{6} + 21 b^{2} ϕ^{5} - 35 b^{3} ϕ^{4} + 35 b^{4} ϕ^{3} - 21 b^{5} ϕ^{2} + (a + 7 b^{6}) ϕ + (b - a b - b^{7})$ , by Theorem 3.2, $7$ does not divide $(Z_{K} : Z [α])$ if and only if $ν_{7} (1 - a - b^{6}) = 1$ .
For $p \notin {2, 3, 7}$ such that $p$ does not divide both $a$ and $b$ , if $p^{2}$ does not divide $6^{6} a^{7} + 7^{7} b^{6}$ , then by the formula $▵ = {(Z_{K} : Z [α])}^{2} d_{K}$ , $p$ does not divide $(Z_{K} : Z [α])$ . If $p^{2}$ divides $6^{6} a^{7} + 7^{7} b^{6}$ , then let $t$ be an integer such that $6 a t \equiv - 7 b (\mod p^{2})$ . Then ${(6 a)}^{6} F^{'} (t) = 7 {(- 7 b)}^{6} + 6^{6} a^{7} \equiv 0 (\mod p^{2})$ and ${(6 a)}^{7} F (t) \equiv 0 (\mod p^{2})$ . Thus ${(x - t)}^{2}$ divides $\bar{F (x)}$ in $F_{p} [x]$ . As $F (t)$ is the remainder of the Euclidean division of $F (x)$ by $x - t$ , by Theorem 3.2, $p$ divides the index $(Z_{K} : Z [α])$ .

For the proofs of Theorems 2.3 and 2.4, we need the following lemma, which characterizes the prime common index divisors of

K

.

Lemma 4.1.

Let

p

be a rational prime integer and

K

be a number field. For every positive integer

f

, let

P_{f}

be the number of distinct prime ideals of

Z_{K}

lying above

p

with residue degree

f

and

N_{f}

the number of monic irreducible polynomials of

F_{p} [x]

of degree

f

. Then

p

is a prime common index divisor of

K

if and only if

P_{f} > N_{f}

for some positive integer

f

.

Proof of Theorem 2.3.

By virtue of Engstrom’s results [14], the proof is done if we provide the factorization of

2 Z_{K}

into powers of prime ideals of

Z_{K}

. Based on Theorem 2.1, we deal with the cases:

2 | a

and

4 | b

or

(a, b) \in {(1, 2), (3, 0)} (\mod 4)

.

If $2$ divides $a$ and $4$ divides $b$ , then for $ϕ = x$ , we have $\bar{F (x)} = ϕ^{7}$ in $F_{2} [x]$ .

(a)

If $N_{ϕ} (F) = S$ has a single side, that is $ν_{2} (a) \geq ν_{2} (b)$ , then the side $S$ is of degree $1$ . Thus there is a unique prime ideal of $Z_{K}$ lying above $2$ .

(b)

If $N_{ϕ} (F) = S_{1} + S_{2}$ has two sides joining $(0, ν_{2} (b))$ , $(1, ν_{2} (a))$ , and $(7, 0)$ , that is $ν_{2} (a) + 1 \leq ν_{2} (b)$ , then $S_{1}$ is of degree $1$ , and so it provides a unique prime ideal of $Z_{K}$ lying above $2$ with residue degree $1$ . Let $d$ be the degree of $S_{2}$ .

i.

If $ν_{2} (a) \notin {2, 3, 4}$ , then $S_{2}$ is of degree $1$ , and so there are exactly two prime ideals of $Z_{K}$ lying above $2$ with residue degree $1$ each.

ii.

If $ν_{2} (a) = 2$ , then the slope of $S_{2}$ is $\frac{- 1}{3}$ and $R_{1} (F) (y) = {(y + 1)}^{2}$ is the residual polynomial of $F (x)$ attached to $S_{2}$ . Thus we have to use second order Newton polygon techniques. Let $ω_{2}$ be the valuation of second order Newton polygon; defined by $ω_{2} (P (x)) = m i n {3 ν_{2} (p_{i}) + i h, ı = 0, \dots, n}$ for every non-zero polynomial $P = \sum_{i = 0}^{n} p_{i} x^{i}$ . Let $ϕ_{2}$ be the key polynomial of $ω_{2}$ and let $N_{2} (F)$ the $ϕ_{2}$ -Newton polygon of $F (x)$ with respect to the valuation $ω_{2}$ . It follows that:

If $ν_{2} (b) = 3$ , then for $ϕ_{2} = x^{3} + 2 x + 2$ , we have $F (x) = x ϕ_{2}^{2} + (4 - 4 x - 4 x^{2}) ϕ_{2} + 8 x^{2} + (a - 4) x + b - 8$ . It follows that if $ν_{2} (a - 4) = 3$ , then $N_{2} (F) = T$ has a single side joining $(0, 10)$ and $(2, 7)$ . Thus $T$ is of degree $1$ , and so $S_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $2$ . If $ν_{2} (a - 4) \geq 4$ and $ν_{2} (b - 8) \geq 4$ , then $N_{2} (F) = T$ has a single side joining $(0, 11)$ , $(1, 9)$ and $(2, 7)$ , with $R_{2} (F) (y) = y^{2} + y + 1$ , which is irreducible over $F_{2} = F_{0}$ . Thus $S_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $2$ with residue degree $2$ . Hence $2$ is not a common index divisor of $K$ .

If $ν_{2} (b) \geq 4$ and $ν_{2} (a + 4) = 3$ , then for $ϕ_{2} = x^{3} + 2$ , we have $F (x) = x ϕ_{2}^{2} - 4 x ϕ_{2} + (a + 4) x + b$ is the $ϕ_{2}$ -expansion of $F (x)$ , and so $N_{2} (F) = T$ has a single side joining $(0, 10)$ and $(2, 7)$ . In this case the side $T$ is of degree $1$ and $S_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $2$ . If $ν_{2} (b) = 4$ and $ν_{2} (a + 4) \geq 4$ , then for $ϕ_{2} = x^{3} + 2$ , $N_{2} (F) = T$ has a single side joining $(0, 12)$ and $(2, 7)$ . Thus $T$ is of degree $1$ , and so $S_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $2$ .

If $ν_{2} (b) \geq 5$ and $ν_{2} (a + 4) = 4$ , then for $ϕ_{2} = x^{3} + 2$ , we have $F (x) = x ϕ_{2}^{2} - 4 x ϕ_{2} + (a + 4) x + b$ is the $ϕ_{2}$ -expansion of $F (x)$ and $N_{2} (F) = T$ has a single side joining $(0, 13)$ , $(1, 10)$ and $(2, 7)$ . So $T$ is of degree $2$ with attached residual polynomial $R_{2} (F) = y^{2} + y + 1$ irreducible over $F_{2} = F_{0}$ . Thus $S_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $2$ with residue degree $2$ .

If $ν_{2} (b) \geq 5$ and $ν_{2} (a + 4) \geq 5$ , then for $ϕ_{2} = x^{3} + 2$ , $N_{2} (F) = T_{1} + T_{2}$ has two sides joining $(0, v)$ , $(1, 10)$ and $(2, 7)$ with $v \geq 15$ . So each $T_{i}$ has degree $1$ , and so $S_{2}$ provides two prime ideals of $Z_{K}$ lying above $2$ with residue degree $1$ each. As $S_{1}$ provides a prime ideal of $Z_{K}$ lying above $2$ with residue degree $1$ , we conclude that there are three prime ideals of $Z_{K}$ lying above $2$ with residue degree $1$ each, and so $2$ is a common index divisor of $K$ . In this last case, $2 Z_{K} = p_{111} p_{121}^{3} p_{131}^{3}$ with residue degree $1$ each prime ideal factor. Based on Engstrom’s result, we conclude that $ν_{2} (i (K)) = 1$ .

iii.

For $ν_{2} (a) = 3$ , we have $R_{1} (F) (y) = y^{3} + 1 = (y^{2} + y + 1) (y + 1)$ is the residual polynomial of $F (x)$ attached to $T_{1}$ . Thus $T_{1}$ provides a unique prime ideal of $Z_{K}$ lying above $2$ , with residue degree $1$ and a unique prime ideal of $Z_{K}$ lying above $2$ with residue degree $2$ . Thus $ν_{2} (i (K)) = 0$ .

iv.

The case $ν_{2} (a) = 4$ is similar to the case $ν_{2} (a) = 2$ . In this case $ν_{2} (i (K)) \geq 1$ if and only if $ν_{2} (b) \geq 7$ and $ν_{2} (a + 16) \geq 7$ . In this case, $2 Z_{K} = p_{111} p_{121}^{3} p_{131}^{3}$ with residue degree $1$ each factor. Based on Engstrom’s result, we conclude that $ν_{2} (i (K)) = 1$ .
$(a, b) \in {(1, 2), (3, 0)} (\mod 4)$ . In this case $\bar{F (x)} = x {(x - 1)}^{2} {(x^{2} + x + 1)}^{2}$ modulo $2$ . Let $ϕ = x - 1$ , $g (x) = x^{2} + x + 1$ , $F (x) = \dots - 21 ϕ^{2} + (7 + a) ϕ + (b + a + 1)$ , and $F (x) = (x - 3) g^{3} + (5 + 3 x) g^{2} - (4 x + 2) g + (a + 1) x + b$ . Since $x$ provides a unique prime ideal of $Z_{K}$ lying above $2$ , we conclude that $2$ is a common index divisor of $K$ if and only if $ϕ$ provides two prime ideals of $Z_{K}$ lying above $2$ of degree $1$ each or $ϕ$ provides a unique prime ideal of $Z_{K}$ lying above $2$ of degree $2$ and $g$ provides at least one prime ideal of $Z_{K}$ lying above $2$ of degree $2$ or also $g$ provides two prime ideals of $Z_{K}$ lying above $2$ of degree $2$ each. That is if and only if one of the following conditions holds:

(a)

If $a \equiv 1 (\mod 4)$ and $b \equiv 2 (\mod 4)$ , then $ν_{2} (▵) \geq 7$ and $N_{g}^{+} (F)$ has a single side of height $1$ , and so $g$ provides a unique prime ideal $p_{311}$ of $Z_{K}$ lying above $2$ with residue degree $2$ . For $N_{ϕ}^{+} (F)$ , let $u = \frac{- 7 b_{2}}{3 a}$ . Then $u \in Z_{2}$ . Let $F (x + u) = x^{7} + \dots + 21 u^{5} x^{2} + A x + B$ , where $A = 7 u^{6} + a = \frac{- ▵}{6^{6} a^{6}}$ and $B = u^{7} + a u + b = \frac{b ▵}{6^{7} a^{7}}$ . It follows that $ν_{2} (A) = ν_{2} (B) = ν_{2} (▵) - 6$ , and so $N_{ϕ}^{+} (F) = S_{1}$ has a single side joining $(0, ν_{2} (▵) - 6)$ and $(2, 0)$ . Thus, if $ν_{2} (▵)$ is odd, then $ϕ$ provides a unique prime ideal $p_{211}$ of $Z_{K}$ lying above $2$ with residue degree $1$ . If $ν_{2} (▵) = 2 (k + 3)$ for some positive integer $k$ , then let $F (x + u + 2^{k}) = x^{7} + \dots + 21 {(u + 2^{k})}^{5} x^{2} + A_{1} x + B_{1}$ , where $A_{1} = 7 u^{6} + a + 3 \cdot 2^{k + 1} u^{5} + 2^{2 k} D = A + 3 \cdot 2^{k + 1} u^{5} + 2^{2 k} D$ and $B_{1} = B + A \cdot 2^{k} + 2^{2 k} \cdot 21 u^{5} + 2^{3 k} H = 2^{2 k} (\frac{b_{2} ▵_{2}}{3^{7} a^{7}} + 21 u^{5}) + 2^{3 k} H$ for some $D \in Z_{2}$ and $H \in Z_{2}$ . Thus, $B_{1} = 2^{2 k} (3 \cdot a \cdot b_{2} ▵_{2} + 3 \cdot a \cdot b_{2}) + 2^{2 k + 3} L$ for some $L \in Z_{2}$ . Hence if $k \geq 2$ , then $ν_{2} (A_{1}) = k + 1$ and $ν_{2} (B_{1}) \geq 2 k + 1$ . More precisely, if $▵_{2} \equiv 1 (\mod 4)$ , then $ν_{2} (B_{1}) = 2 k + 1$ , and so $ϕ$ provides a unique prime ideal $p_{211}$ of $Z_{K}$ lying above $2$ with residue degree $1$ . If $▵_{2} \equiv 3 (\mod 4)$ , then $ν_{2} (B_{1}) \geq 2 k + 2$ . It follows that if $ν_{2} (B_{1}) = 2 k + 2$ , and so $ϕ$ provides a unique prime ideal $p_{211}$ of $Z_{K}$ lying above $2$ with residue degree $2$ . If $ν_{2} (B_{1}) \geq 2 k + 3$ , then $ν_{2} (B_{1}) \geq 2 k + 3$ , and so $ϕ$ provides two prime ideals of $Z_{K}$ lying above $2$ with residue degree $1$ each. In these last two cases, we have $2$ divides $i (K)$ and $ν_{2} (i (K)) = 1$ .

For $k = 1$ , we have $ν_{2} (▵) = 8$ and $a \equiv 5 (\mod 8)$ . In this case $\bar{F (x)} = x {(x - 1)}^{2} {(x^{2} + x + 1)}^{2}$ modulo $2$ . Let $ϕ = x - 1$ , $g (x) = x^{2} + x + 1$ , $F (x) = \dots - 21 ϕ^{2} + (7 + a) ϕ + (b + a + 1)$ , and $F (x) = (x - 3) g^{3} + (5 + 3 x) g^{2} - (4 x + 2) g + (a + 1) x + b$ . Since $x$ provides a unique prime ideal of of $Z_{K}$ lying above $2$ with residue degree $1$ and $g$ provides a unique prime ideal of of $Z_{K}$ lying above $2$ with residue degree $2$ , we conclude that $ν_{2} (i (K)) \geq 1$ if and only if $ϕ$ provides a unique prime ideal of of $Z_{K}$ lying above $2$ with residue degree $2$ or $ϕ$ provides two distinct prime ideals of of $Z_{K}$ lying above $2$ with residue degree $1$ each. If $(a, b) \in {(5, 10), (13, 2)} (\mod 16)$ , then $ϕ$ provides a unique prime ideal of $Z_{K}$ lying above $2$ with residue degree $2$ and so $ν_{2} (i (K)) = 1$ . If $(a, b) \in {(5, 10), (13, 10)} (\mod 16)$ , then $ϕ$ provides a unique prime ideal of $Z_{K}$ lying above $2$ with residue degree $1$ and so $ν_{2} (i (K)) = 0$ . For $(a, b) \in {(5, 6), (5, 14), (13, 6), (13, 14)} (\mod 16)$ , let us replace $ϕ$ by $ϕ^{'} = x - 3$ and consider the $ϕ^{'}$ -Newton polygon of $F (x)$ with respect to $ν_{2}$ . It follows that If $(a, b) \in {(5, 6), (13, 14)} (\mod 16)$ , then $ϕ^{'}$ provides two prime ideals of $Z_{K}$ lying above $2$ with residue degree $1$ each and so $ν_{2} (i (K)) = 1$ . If $(a, b) \in {(5, 14), (13, 6)} (\mod 16)$ , then $ϕ^{'}$ provides a unique prime ideal of $Z_{K}$ lying above $2$ with residue degree $1$ and so $ν_{2} (i (K)) = 0$ .

(b)

$a \equiv 3 (\mod 4)$ and $b \equiv - (a + 1) (\mod 8)$ because $N_{ϕ}^{+} (F)$ has two sides.

(c)

If $a \equiv 3 (\mod 8)$ and $b \equiv 0 (\mod 8)$ , then $ϕ$ provides a unique prime ideal of $Z_{K}$ lying above $2$ with residue degree $2$ and $g$ provides two prime ideals of $Z_{K}$ lying above $2$ with residue degree $2$ each because $N_{g}^{+} (F)$ has a single side of degree $2$ with $(1 + x) y^{2} + y + x = (x + 1) (y - 1) (y - x)$ its attached residual polynomial of $F (x)$ . In this case $2 Z_{K} = p_{111} p_{211} p_{311} p_{312}$ with residue degrees $f_{111} = 1$ and $f_{211} = f_{311} = f_{312} = 2$ , and so $ν_{2} (i (K)) = 3$ .

(d)

$a \equiv 7 (\mod 8)$ and $b \equiv 0 (\mod 8)$ . In this case $ϕ$ provides a unique prime ideal of $Z_{K}$ lying above $2$ with residue degree $2$ and $N_{g}^{+} (F)$ has two sides. More precisely, $2 Z_{K} = p_{111} p_{211} p_{311} p_{321}$ with residue degrees $f_{111} = 1$ and $f_{211} = f_{311} = f_{312} = 2$ , and so $ν_{2} (i (K)) = 3$ .

(e)

If $a \equiv 5 (\mod 8)$ and $b \equiv - (a + 1) (\mod 16)$ because if $b \equiv - (a + 1) (\mod 32)$ , then $N_{ϕ}^{+} (F)$ has two sides and if $b \equiv - (a + 1) + 16 (\mod 32)$ , then $N_{ϕ}^{+} (F)$ has a single side of degree $2$ , which provides a single prime ideal of $Z_{K}$ lying above $2$ with residue degree $2$ and $N_{g}^{+} (F)$ has a single side of degree $1$ . Thus there are $2$ prime ideals of $Z_{K}$ lying above $2$ with residue degree $2$ each.

(f)

If $a \equiv 5 (\mod 8)$ and $ν_{2} (b + (a + 1)) = 2$ , then $ν_{2} (b - (a + 1)) \geq 3$ . If $ν_{2} (b - (a + 1)) = 3$ , then for $ϕ = x + 1$ , we have $N_{ϕ}^{+} (F)$ has a single side of degree $1$ . Since $ν_{2} (a + 1) = 1$ , then $N_{g}^{+} (F)$ has a single side of height $1$ . Thus there are two prime ideals of $Z_{K}$ lying above $2$ with residue degree $1$ each and one prime ideal with residue degree $2$ . If $ν_{2} (b - (a + 1)) = 4$ , then for $ϕ = x + 1$ , we have $N_{ϕ}^{+} (F)$ has a single side of degree $2$ and its attached residual polynomial of $F$ is $R_{1} (F) (y) = y^{2} + y + 1$ . Since $b \equiv 6 (\mod 8)$ , we conclude that $N_{g}^{+} (F)$ has a single side of degree $1$ , then there are $2$ prime ideals of $Z_{K}$ lying above $2$ with residue degree $2$ each, and so $2$ divides $i (K)$ . If $ν_{2} (b - (a + 1)) \geq 5$ , then for $ϕ = x + 1$ , we have $N_{ϕ}^{+} (F)$ has two sides of degree $1$ each, and so there are $3$ prime ideals of $Z_{K}$ lying above $2$ with residue degree $1$ each, and so $2$ divides $i (K)$ .

Proof of Theorem 2.4.

By virtue of Engstrom’s results [14], the proof is done if we provide the factorization of

3 Z_{K}

into powers of prime ideals of

Z_{K}

. Based on Theorem 2.1, we deal with the cases:

$3 | a$ and $9 | b$ .
$(a, b) \notin {(2, 0), (8, 3), (8, 6)} (\mod 9)$ .
$(a, b) \equiv (1, 0) (\mod 9)$ .

$3 | a$ and $9 | b$ , then for $ϕ = x$ , $\bar{F (x)} = ϕ^{7}$ in $F_{3} [x]$ . It follows that:

(a)

If $ν_{3} (a) \geq ν_{3} (b)$ , then $N_{ϕ} (F)$ has a single side of degree $1$ , and so there is a unique prime ideal of $Z_{K}$ lying above $3$ .

(b)

If $ν_{3} (a) + 1 \leq ν_{3} (b)$ , then $N_{ϕ} (F) = S_{1} + S_{2}$ has two sides joining $(0, ν_{3} (b))$ , $(1, ν_{3} (a))$ , and $(7, 0)$ . Since $S_{1}$ is of degree $1$ , $S_{1}$ provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ . Thus $ν_{3} (i (K)) \geq 1$ if and only if $S_{2}$ provides at least three prime ideals of $Z_{K}$ lying above $3$ , with residue degree $1$ each. If $ν_{3} (a) \notin {2, 3, 4}$ , then $S_{2}$ is of degree $1$ , and so $S_{2}$ provides exactly one prime ideal of $Z_{K}$ lying above $3$ , with residue degree $1$ each. If $ν_{3} (a) \in {2, 4}$ , then $S_{2}$ is of degree $2$ , and so $S_{2}$ provides at most two prime ideal of $Z_{K}$ lying above $3$ . Hence $3$ is not a common index divisor of $K$ . If $ν_{3} (a) = 3$ , then $S_{2}$ is of degree $3$ and its attached residual polynomial of $F (x)$ is $R_{1} (F) (y) = y^{3} + a_{3} = {(y + a_{3})}^{3}$ . So, we have to use second order Newton polygon. Let $ω_{2}$ be the valuation of second order Newton polygon. $ω_{2}$ is defined by $ω_{2} (P) = m i n {2 ν_{3} (p_{i}) + i, i = 0, \dots, n}$ for every non zero polynomial $p = \sum_{i = 0}^{n} p_{i} x^{i}$ of $Q_{3} [x]$ . Let $ϕ_{2} = x^{2} + 3 a_{3}$ be a key polynomial of $ω_{2}$ and $N_{2} (F)$ the $ϕ_{2}$ -Newton polygon of $F (x)$ with respect to $ω_{2}$ . It follows that: If $a_{3} \equiv 1 (\mod 3)$ , then for $ϕ_{2} = x^{2} + 3$ , we have $F (x) = x ϕ_{2}^{3} - 9 x ϕ_{2}^{2} + 27 x ϕ_{2} + (a - 27) x + b$ is the $ϕ_{2}$ -expansion of $F (x)$ . We have the following cases:

i.

If $ν_{3} (b) = 4$ , then $N_{2} (F) = T$ has a single side joining $(0, 8)$ and $(3, 7)$ . Thus $T$ is of degree $1$ and $S_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ .

ii.

If $ν_{3} (b) \geq 5$ and $ν_{3} (a - 27) = 4$ , then $N_{2} (F) = T$ has a single side joining $(0, 9)$ and $(3, 7)$ . Thus $T$ is of degree $1$ and $S_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ .

iii.

If $ν_{3} (b) = 5$ and $ν_{3} (a - 27) \geq 5$ , then $N_{2} (F) = T$ has a single side joining $(0, 10)$ and $(3, 7)$ and its attached residual polynomial of $F$ is $R_{2} (F) (y) = x y^{3} + x y + b_{3}$ , which is irreducible over $F_{2} = F_{ϕ}$ because $ϕ$ is of degree $1$ . Thus $S_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $3$ .

iv.

If $ν_{3} (b) \geq 6$ and $ν_{3} (a - 27) \geq 5$ , then $N_{2} (F) = T_{1} + T_{2}$ has two sides joining $(0, v)$ , $(2, 9)$ and $(3, 7)$ with $v \geq 11$ . Thus $T_{1}$ is of degree $1$ , $T_{2}$ of degree $2$ and $R_{2} (F) (y) = x y^{2} + x$ is its attached residual polynomial of $F (x)$ , which is irreducible over $F_{2} = F_{ϕ}$ . Thus $S_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $3$ , with residue degree $1$ and a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $2$ .

Similarly, for $a_{3} \equiv - 1 (\mod 3)$ , let $ϕ_{2} = x^{2} - 3$ . Then $F (x) = x ϕ_{2}^{3} + 9 x ϕ_{2}^{2} + 27 x ϕ_{2} + (a + 27) x + b$ is the $ϕ_{2}$ -expansion of $F (x)$ . By analogous to the case $a_{3} \equiv 1 (\mod 3)$ , in every case $3$ does not divide $i (K)$ . If $a \equiv 1 (\mod 3)$ , then $\bar{F (x)} = x {(x^{2} + 1)}^{3}$ in $F_{3} [x]$ . So, there are exactly a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ and the other prime ideals of $Z_{K}$ lying above $3$ are of residue degrees at least $2$ each prime ideal factor. Hence $ν_{3} (i (K)) = 0$ .

(c)

If $a \equiv - 1 (\mod 3)$ , then $\bar{F (x)} = x {(x - 1)}^{3} {(x + 1)}^{3}$ in $F_{3} [x]$ . Let $ϕ_{1} = x - 1$ , $ϕ_{2} = x + 1$ , $F (x) = ϕ_{1}^{7} + 7 ϕ_{1}^{6} + 21 ϕ_{1}^{5} + 35 ϕ_{1}^{4} + 35 ϕ_{1}^{3} + 21 ϕ_{1}^{2} + (7 + a) ϕ_{1} + (b + a + 1)$ , and $F (x) = ϕ_{2}^{7} - 7 ϕ_{2}^{6} + 21 ϕ_{2}^{5} - 35 ϕ_{2}^{4} + 35 ϕ_{2}^{3} - 21 ϕ_{2}^{2} + (7 + a) ϕ_{2} + (b - (a + 1))$ . It follows that:

i.

If $a \equiv 8 (\mod 9)$ and $b \equiv 0 (\mod 9)$ , then $ν_{3} (b + (1 + a)) \geq 2$ and $ν_{3} (b - (1 + a)) \geq 2$ . Thus $x$ a provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ , and each $ϕ_{i}$ provides two prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each prime ideal factor. In this two cases $ν_{3} (i (K)) = 2$ .

ii.

If $a \equiv 5 (\mod 9)$ and $b \equiv 3 (\mod 9)$ , then $ν_{3} (b + (1 + a)) \geq 2$ and $ν_{3} (b - (1 + a)) = 1$ . Thus each of $x$ and $ϕ_{2}$ provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ , and $ϕ_{1}$ provides two prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each. Similarly, if $a \equiv 5 (\mod 9)$ and $b \equiv 6 (\mod 9)$ , then $ν_{3} (b - (1 + a)) \geq 2$ and $ν_{3} (b + (1 + a)) = 1$ . Thus each of $x$ and $ϕ_{1}$ provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ , $ϕ_{2}$ provides two prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each. In these two cases $ν_{3} (i (K)) = 1$ .

iii.

If $a \equiv 2 (\mod 9)$ and $b \equiv (1 + a) \pm 9 (\mod 27)$ , then $N_{ϕ_{2}}^{+} (F)$ has a single side joining $(0, 2)$ and $(3, 0)$ and $N_{ϕ_{1}}^{+} (F)$ has a single side joining $(0, 1)$ and $(3, 0)$ . Thus there are $3$ prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each, and so $ν_{3} (i (K)) = 0$ .

iv.

Similarly, if $a \equiv 2 (\mod 9)$ and $b \equiv - (1 + a) \pm 9 (\mod 27)$ , then there are $3$ prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each, and so $ν_{3} (i (K)) = 0$ .

v.

If $a \equiv 2 (\mod 9)$ and $ν_{3} (b) = 1$ , then $ν_{3} (▵) \geq 8$ . Let $u = \frac{- 7 b_{3}}{2 a}$ . Then $u \in Z_{3}$ . Let $ϕ = x - u$ and $F (x + u) = x^{7} + \dots + 35 u^{4} x^{3} + 21 u^{5} x^{2} + A x + B$ , where $A = 7 u^{6} + a = \frac{- ▵}{6^{6} a^{6}}$ and $B = u^{7} + a u + b = \frac{b ▵}{6^{7} a^{7}}$ . It follows that $ν_{3} (A) = ν_{3} (B) = ν_{3} (▵) - 6$ , and so $N_{ϕ}^{+} (F) = S_{1}$ has a single side joining $(0, ν_{3} (▵) - 6)$ and $(3, 0)$ . Remark that since $ν_{3} (b) = 1$ and $ν_{3} (B) \geq 2$ , $ν_{3} (- u^{7} - a u + b) = 1$ , and so $(x + u)$ provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ . Thus $ν_{3} (i (K)) \geq 1$ if and only if $ϕ$ provides at least two prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each prime ideal factor.

A.

If $ν_{3} (▵) = 8$ , then $N_{ϕ}^{+} (F)$ has a single side of degree one, and so $ϕ$ provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ .

B.

If $ν_{3} (▵) = 9$ , then $N_{ϕ}^{+} (F) = S$ has a single side joining $(0, 3)$ and $(3, 0)$ with $R_{1} (F) (y) = - u^{4} y^{3} + u^{5} y^{2} + B_{3}$ its attached residual polynomial of $F (x)$ . Since $a \equiv - 1 (\mod 3)$ and $B = \frac{b ▵}{6^{7} a^{7}}$ , we have $u \equiv - b_{3} (\mod 3)$ and $B_{3} \equiv b_{3} ▵_{3} (\mod 3)$ . Thus $R_{1} (F) (y) = - y^{3} - b_{3} y^{2} + b_{3} ▵_{3}$ . Since $R_{1} (F) (y)$ is square free and $R_{1} (F) (0) \neq 0$ , then $R_{1} (F) (y)$ has at most one root in $F_{ϕ}$ . Thus $S$ provides at most a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ . Therefore, $ν_{3} (i (K)) = 0$ .

C.

If $ν_{3} (▵) \geq 10$ , then $N_{ϕ}^{+} (F) = S_{1} + S_{2}$ has two sides joining $(0, v - 6)$ and $(3, 1)$ . It follows that Since $S_{2}$ is of degree $1$ , it provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $1$ . Moreover, if $ν_{3} (▵)$ is even then $S_{1}$ is of degree $1$ , and so $ϕ$ provides two prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each. In this case $ν_{3} (i (K)) = 1$ . If $ν_{3} (▵) = 2 (k + 3) + 1$ , then $S_{1}$ is of degree $2$ with residual polynomial $R_{1} (F) (y) = u y^{2} + b_{3} ▵_{3}$ . Since $a \equiv - 1 (\mod 3)$ , we have $2 a \equiv 1 (\mod 3)$ and $u \equiv - b_{3} (\mod 3)$ . Thus $R_{1} (F) (y) = - b_{3} (y^{2} - ▵_{3})$ . It follows that if $(\frac{▵_{3}}{3}) = 1$ , then $R_{1} (F) (y)$ has two different factors of degree $1$ each, and so $S_{1}$ provides two prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each. In this case there are exactly five prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each and according to Engstrom’s results $ν_{3} (i (K)) = 2$ . But if $(\frac{▵_{3}}{3}) = - 1$ , then $R_{1} (F) (y)$ is irreducible over $F_{ϕ} = F_{3}$ , and so $S_{1}$ provides a unique prime ideal of $Z_{K}$ lying above $3$ with residue degree $2$ . In this last case there are exactly three prime ideals of $Z_{K}$ lying above $3$ with residue degree $1$ each, and so $ν_{3} (i (K)) = 0$ .

Proof of Theorem 2.5.

We start by showing that

5

does not divide

i (K)

for every integers of

a

and

b

such that

x^{7} + a x + b

is irreducible. By virtue of Engstrom’s results [14], the proof is done if we provide the factorization of

5 Z_{K}

into powers of prime ideals of

Z_{K}

. By by the index formula

▵ = {(Z_{K} : Z [α])}^{2} d_{K}

, if

5^{2}

does not divide

▵

, then

ν_{5} (i (K)) = 0

. So, we assume that

5^{2}

divides

▵

.

So, $6^{6} a^{7} + 7^{7} b^{6} \equiv 0 (\mod 5)$ . Since $a^{5} \equiv a (\mod 5)$ and $b^{5} \equiv b (\mod 5)$ , then $a^{3} \equiv 2 b^{2} (\mod 5)$ , which means $(a, b) \in {(0, 0), (3, 1), (2, 2), (2, 3), (3, 4)} (\mod 5)$ . In order to show that $ν_{5} (i (K)) = 0$ it suffices to show that for every value $(a, b) \in Z^{2}$ such that $x^{7} + a x + b$ is irreducible and $(a, b) \in {(0, 0), (2, 1), (3, 2), (3, 3), (2, 34} (\mod 5)$ there are at most four prime ideals of $Z_{K}$ lying above $5$ with residue degree $1$ , where $K$ is the number field generated by a complex root of $x^{7} + a x + b$ .

(a)

For $(a, b) \equiv (0, 0) (\mod 5)$ , if $ν_{5} (a) \geq ν_{5} (b)$ , then $N_{ϕ} (F) = S$ has a single side and it is of degree $1$ . Thus there is a unique prime ideal $p$ of $Z_{K}$ lying above $5$ with residue degree $1$ . More precisely $5 Z_{K} = p^{7}$ .

If $ν_{5} (a) + 1 \leq ν_{5} (b)$ , then $N_{ϕ} (F) = S_{1} + S_{2}$ has two sides. More precisely, $S_{1}$ is of degree $1$ . Let $d$ be degree of $S_{2}$ . Since $6$ is the length of $S_{2}$ , then $d \in {1, 2, 3}$ . Thus $S_{1}$ provides a unique prime ideal $p$ of $Z_{K}$ lying above $5$ with residue degree $1$ and $S_{2}$ provides at most three prime ideals $p$ of $Z_{K}$ lying above $5$ with residue degree $1$ each.

(b)

For $(a, b) \equiv (3, 1) (\mod 5)$ , since $\bar{F (x)} = {(x + 4)}^{2} (x + 3) (x^{4} + 4 x^{3} + x^{2} + x + 2)$ in $F_{5} [x]$ , there are at most three prime ideals $p$ of $Z_{K}$ lying above $5$ with residue degree $1$ each.

(c)

For $(a, b) \equiv (2, 2) (\mod 5)$ , since $\bar{F (x)} = (x^{4} + 2 x^{3} + 4 x^{2} + 2 x + 2) (x + 4) {(x + 2)}^{2}$ in $F_{5} [x]$ , there are at most three prime ideals $p$ of $Z_{K}$ lying above $5$ with residue degree $1$ each.

(d)

For $(a, b) \equiv (2, 3) (\mod 5)$ , since $\bar{F (x)} = (x + 1) {(x + 3)}^{2} (x^{4} + 3 x^{3} + 4 x^{2} + 3 x + 2)$ in $F_{5} [x]$ , there are at most three prime ideals $p$ of $Z_{K}$ lying above $5$ with residue degree $1$ each.

(e)

For $(a, b) \equiv (3, 4) (\mod 5)$ , since $\bar{F (x)} = (x^{4} + x^{3} + x^{2} + 4 x + 2) {(x + 1)}^{2} (x + 2)$ in $F_{5} [x]$ , there are at most three prime ideals $p$ of $Z_{K}$ lying above $5$ with residue degree $1$ each.

We conclude that in all cases

ν_{5} (i (K)) = 0

.

For

p \geq 7

, since the field

K

is of degree

7

, there are at most

7

prime ideals of

Z_{K}

lying above

p

. The fact that there at least

p \geq 7

monic irreducible polynomial of degree

f

in

F_{p} [x]

for every positive integer

f \in {1, 2, 3}

, we conclude that

p

does not divide

i (K)

.

Proof of Proposition 2.2.

First according to Theorem 2.1 and the hypotheses of Example 2.2,

2

is the unique prime integer candidate to divide

i n d (α)

. Let

ϕ = x

. Then

\bar{F (x)} = ϕ^{7}

in

F_{2} [x]

and

N_{ϕ} (F) = S

has a single side of degree GCD

(7, v) = 1

. Thus

F (x)

is irreducible over

Q_{2}

. Let

K

be the number field generated by a root

α

of

F (x)

. Since

F (x)

is irreducible over

Q_{2}

, there is a unique valuation

ω

of

K

extending

ν_{2}

. By Theorem 3.1, we have

ν_{2} (Z_{K} : Z [α]) \geq i n d_{ϕ} (F) \geq 1

, and so

F (x)

is not a monogenic polynomial. Let

θ = \frac{α^{x}}{2^{y}}

, where

(x, y)

is the unique solution of integers of the Diophantine equation

v x - 7 y = 1

and

0 \leq x \leq 6

. Then

θ \in K

. Since

v

and

7

are coprime, we conclude that

K = Q (θ)

. Let us show that

Z_{K} = Z [θ]

, and so

K

is monogenic. By [13, Corollary 3.1.4], in order to show that

θ \in Z_{K}

, we need to show that

ω (θ) \geq 0

, where

ω

is the unique valuation of

K

extending

ν_{2}

. Since

N_{ϕ} (F) = S

has a single side of slope

- v / 7

, we conclude that

ω (α) = v / 7

, and so

ω (θ) = \frac{x v}{7} - y = \frac{1}{7}

. Let

g (x)

be the minimal polynomial of

θ

over

Q

. By the formula relating roots and coefficients of a monic polynomial, we conclude that

g (x) = x^{7} + \sum_{i = 1}^{7} {(- 1)}^{i} s_{i} x^{7 - i}

, where

s_{i} = \sum_{k_{1} < \dots < k_{i}} θ_{k_{1}} \dots θ_{k_{i}}

and

θ_{1}, \dots, θ_{7}

are the

Q_{p}

-conjugates of

θ

. Since there is a unique valuation extending

ν_{2}

to any algebraic extension of

Q_{2}

, we conclude that

ω (θ_{i}) = 1 / 7

for every

i = 1, \dots, 7

. Thus

ν_{2} (s_{7}) = ω (θ_{1} \dots θ_{7}) = 7 \times 1 / 7 = 1

and

ν_{2} (s_{i}) \geq i / 7

for every

i = 1, \dots, 6

, which means that

g (x)

is a

2

-Eisenstein polynomial. Hence

2

does not divide the index

(Z_{K} : Z [θ])

. As

2

is the unique positive prime integer candidate to divide

(Z [α] : Z [θ])

, we conclude that for every prime integer

p

,

p

does not divide

(Z_{K} : Z [θ])

, which means that

Z_{K} = Z [θ]

.

5. Examples

Let

F = x^{7} + a x + b \in Z [x]

be a monic irreducible polynomial and

K

a number field generated by a root

α

of

F (x)

. In the following examples, we calculate the index of the field

K

. First based on Theorem 2.5,

ν_{p} (i (K)) = 0

for every prime integer

p \geq 5

. Thus we need only to calculate

ν_{p} (i (K))

for

p = 2, 3

.

For $a = 6$ and $b = 6$ , since $F (x)$ is $p$ -Eisenstein for every $p = 2, 3$ , we conclude that $F (x)$ is irreducible over $Q$ , $2$ (resp. $3$ ) does not divide $(Z_{K} : Z [α])$ . Thus $2$ (resp. $3$ ) does not divide $i (K))$ , and so $i (K) = 1$ .
For $a = 28$ and $b = 32$ , since $\bar{F (x)}$ is irreducible over $F_{5}$ , $F (x)$ is irreducible over $Q$ . By the first item of Theorem 2.3, we have $ν_{2} (i (K)) = 1$ . By Theorem 2.4, $ν_{3} (i (K)) = 0$ . Thus $i (K) = 2$ .
For $a = 3$ and $b = 8$ , $\bar{F (x)}$ is irreducible over $F_{5}$ , $F (x)$ is irreducible over $Q$ . Again since $a \equiv 3 (\mod 4)$ and $b \equiv 0 (\mod 8)$ , by Theorem 2.3, $ν_{2} (i (K)) = 3$ . By Theorem 2.4, $ν_{3} (i (K)) = 0$ . Thus $i (K) = 8$ .
For $a = - 1$ and $b = 9$ , since $\bar{F (x)}$ is irreducible over $F_{2}$ , $F (x)$ is irreducible over $Q$ . Since $2 Z_{K}$ is a prime ideal of $Z_{K}$ , $ν_{2} (i (K)) = 0$ . Also since $a \equiv 8 (\mod 9)$ and $b \equiv 0 (\mod 9)$ , by Theorem 2.4, $ν_{3} (i (K)) = 2$ . Thus $i (K) = 9$ .
For $a = 803$ and $b = 2112$ , since $\bar{F (x)}$ is irreducible over $F_{5}$ , $F (x)$ is irreducible over $Q$ . Since $a \equiv 3 (\mod 4)$ and $b \equiv 0 (\mod 8)$ , by Theorem 2.3, $ν_{2} (i (K)) = 3$ . Similarly since $a \equiv 5 (\mod 9)$ and $b \equiv 6 (\mod 9)$ , by Theorem 2.4, $ν_{3} (i (K)) = 1$ . Thus $i (K) = 24$ .
For $a = 35$ and $b = 72$ , since $\bar{F (x)}$ is irreducible over $F_{11}$ , $F (x)$ is irreducible over $Q$ . Since $a \equiv 3 (\mod 4)$ and $b \equiv 0 (\mod 8)$ , by Theorem 2.3, $ν_{2} (i (K)) = 3$ . Similarly since $a \equiv 8 (\mod 9)$ and $b \equiv 0 (\mod 9)$ , by Theorem 2.4, $ν_{3} (i (K)) = 2$ . Thus $i (K) = 72$ .

Conflicts of Interest

There are non-financial competing interests to report.

Data Availability Statement

Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

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